Answer:
At a certain pizza parlor,36 % of the customers order a pizza containing onions,35 % of the customers order a pizza containing sausage, and 66% order a pizza containing onions or sausage (or both). Find the probability that a customer chosen at random will order a pizza containing both onions and sausage.
Step-by-step explanation:
Hello!
You have the following possible pizza orders:
Onion ⇒ P(on)= 0.36
Sausage ⇒ P(sa)= 0.35
Onions and Sausages ⇒ P(on∪sa)= 0.66
The events "onion" and "sausage" are not mutually exclusive, since you can order a pizza with both toppings.
If two events are not mutually exclusive, you know that:
P(A∪B)= P(A)+P(B)-P(A∩B)
Using the given information you can use that property to calculate the probability of a customer ordering a pizza with onions and sausage:
P(on∪sa)= P(on)+P(sa)-P(on∩sa)
P(on∪sa)+P(on∩sa)= P(on)+P(sa)
P(on∩sa)= P(on)+P(sa)-P(on∪sa)
P(on∩sa)= 0.36+0.35-0.66= 0.05
I hope it helps!
<span>Assuming that Josh throws the discus 36 times, the expected value of y will be the length each discus reaches for each throw all added together. Specifically, it will be discus throw one, plus discus throw two -- all the way to discus throw thirty-six. Adding all of those throws together will provide the expected value of Y.</span>
Answer:
Step-by-step explanation:
Product of 5 and a number: 5n
Six more than that would be 5n + 6
Finally, "six more than the product of 5 and a number, decreased by one" would be
5n + 6 - 1, or 5n + 5
Perimeter is the sum of all the sides of the shape
do you know the length of each side of the parallelogram?
